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      <title>如何使用NIO和JUC提升性能 - 交互式测验</title>
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          <div class="quiz-container">
            <div class="quiz-header">
              <h1>如何使用NIO和JUC提升性能 - 交互式测验</h1>
              <p>检验你对文档内容的理解程度</p>
            </div>
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      <script>
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  {
    "question": "根据文档，NIO（非阻塞I/O）相较于传统BIO（阻塞I/O）在处理网络连接时的核心优势是什么？",
    "options": [
      "每个连接都必须独占一个线程，保证了连接的稳定性。",
      "一个线程可以处理多个连接，提高了线程利用率和系统的可扩展性。",
      "NIO完全不需要线程，通过硬件中断直接处理I/O。",
      "NIO通过增加线程池大小来处理更多连接，而BIO不能使用线程池。"
    ],
    "correctAnswerIndex": 1,
    "explanation": "文档中明确指出，NIO的线程模型是“1线程处理多连接”，而BIO是“1线程处理1连接”。这使得NIO在高并发场景下具有优秀的可扩展性，因为它避免了为每个连接都创建一个线程所带来的巨大资源开销和上下文切换成本。",
    "source": "文档中「2.1 传统BIO vs NIO」部分",
    "hint": "回顾一下文档中对比BIO和NIO的表格，特别关注“线程模型”这一行。"
  },
  {
    "question": "在Java NIO编程模型中，哪个核心组件扮演着“多路复用器”的角色，能够同时监听多个Channel上的事件（如连接、读、写）？",
    "options": [
      "Buffer",
      "Channel",
      "Selector",
      "ThreadPool"
    ],
    "correctAnswerIndex": 2,
    "explanation": "Selector是NIO实现非阻塞和高并发的关键。它允许单个线程监视多个Channel的文件描述符。当某个Channel准备好进行I/O操作时，Selector会得到通知，从而实现高效的事件驱动模型，避免了线程的忙等待或为每个连接创建线程。",
    "source": "文档中「2.2 NIO核心组件」部分",
    "hint": "文档中提到了三个NIO核心组件，其中一个被描述为“多路复用器，监听多个Channel事件”。"
  },
  {
    "question": "在“NIO + JUC实战：高性能HTTP服务器”的架构设计中，NIO和JUC线程池分别承担了什么核心职责？",
    "options": [
      "NIO负责处理网络I/O事件（如连接接收和数据读写），而JUC线程池负责执行耗时的业务逻辑。",
      "JUC线程池负责监听网络端口，NIO负责处理具体的业务计算。",
      "NIO和JUC线程池共同负责网络I/O，业务逻辑由主线程处理。",
      "NIO只负责建立连接，所有的数据读写和业务逻辑都由JUC线程池处理。"
    ],
    "correctAnswerIndex": 0,
    "explanation": "这是一种典型的Reactor模式应用。I/O线程（由NIO的Selector驱动）专门负责快速响应网络事件，并将解码后的业务请求分发给后端的业务线程池（由JUC提供）处理。这种分工避免了耗时的业务逻辑阻塞宝贵的I/O线程，从而最大化系统的吞吐量。",
    "source": "文档中「4.1 架构设计」部分",
    "hint": "在实战的架构设计中，第一点和第二点明确地划分了这两者的工作范围。"
  },
  {
    "question": "文档提到了多种JUC并发集合，其中哪一个特别适用于“读多写少”的场景，通过写入时复制的策略来保证线程安全？",
    "options": [
      "ConcurrentHashMap",
      "CopyOnWriteArrayList",
      "BlockingQueue",
      "AtomicInteger"
    ],
    "correctAnswerIndex": 1,
    "explanation": "CopyOnWriteArrayList在修改（add, set, remove等）操作时，会创建一个底层数组的副本来进行操作，完成后再将引用指向新的数组。这个过程是加锁的。而读操作不加锁，直接访问旧的数组引用，因此非常适合读操作远多于写操作的并发场景。",
    "source": "文档中「3.2 并发集合」部分",
    "hint": "想一想哪个集合的名称本身就暗示了它的工作机制——“写入时复制”。"
  },
  {
    "question": "根据“性能对比”表格，从BIO模型切换到NIO+JUC模型后，系统表现出以下哪项显著提升？",
    "options": [
      "线程数大幅增加，但降低了CPU利用率以节省能耗。",
      "内存占用更高，但换来了更低的平均响应时间。",
      "使用更少的线程，实现了更高的CPU利用率和QPS（每秒查询率）。",
      "QPS基本不变，但系统的线程数和内存占用都显著降低。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "对比表格清晰地展示了NIO+JUC模型的优势：线程数从1000降至16，CPU利用率从30%提升到80%，内存占用从高降低，最大QPS从1200飙升至9800，同时平均响应时间也大大缩短。这体现了其高效的资源利用能力。",
    "source": "文档中「五、性能对比 传统BIO vs NIO+JUC」部分",
    "hint": "仔细查看对比表格中的每一项指标，看看NIO+JUC相比BIO是变好了还是变差了。"
  },
  {
    "question": "在“性能优化点”一节中，提到的“零拷贝技术”主要是为了解决什么问题？",
    "options": [
      "减少代码的重复拷贝，提高代码复用性。",
      "实现对象的无拷贝创建，降低GC压力。",
      "减少数据在内核空间和用户空间之间的内存拷贝次数，提升数据传输效率。",
      "避免在多线程之间拷贝数据，通过共享内存通信。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "传统的文件传输需要将数据从内核缓冲区拷贝到用户缓冲区，再从用户缓冲区拷贝到Socket的内核缓冲区。零拷贝技术（如`FileChannel.transferTo`）允许操作系统直接将数据从文件系统的内核缓冲区发送到Socket的内核缓冲区，省去了中间的用户空间拷贝环节，从而显著减少了CPU消耗和内存带宽占用。",
    "source": "文档中「4.3 性能优化点」部分",
    "hint": "思考一下“拷贝”这个词在I/O上下文中通常指的是什么？优化点提到了一个具体的Java类和方法来实现它。"
  },
  {
    "question": "文档在“进阶话题”中提到，像Netty这样的高性能网络框架，通常采用哪种高效的Reactor模式设计？",
    "options": [
      "单Reactor单线程模式",
      "主从Reactor多线程模式",
      "单Reactor多线程模式",
      "ForkJoinPool模式"
    ],
    "correctAnswerIndex": 1,
    "explanation": "主从Reactor多线程模式是Netty等高性能网络框架的基石。它使用一个主Reactor专门处理连接请求，然后将建立好的连接分发给多个从Reactor。每个从Reactor负责处理数据读写和业务逻辑，实现了职责分离和负载均衡，极大地提升了系统的并发处理能力。",
    "source": "文档中「6.1 Reactor模式」部分",
    "hint": "文档中列出了三种Reactor模式，并明确指出了Netty采用的是哪一种。"
  }
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                '<div class="report-actions">' +
                '<button id="review-btn">回顾测验</button>' +
                '<button class="primary" onclick="location.reload()">再试一次</button>' +
                '</div>' +
                '</div>';

            let reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }
            quizContentEl.insertAdjacentHTML('beforeend', finalScoreHtml);
            document.getElementById('review-btn').addEventListener('click', reviewQuiz);
        }

        function reviewQuiz() {
            const reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }

            document.querySelector('.quiz-body').style.display = 'block';
            document.getElementById('quiz-footer').style.display = 'flex';
            
            currentQuestionIndex = 0;
            loadQuestion();
            updateNavigationButtons(); // Ensure nav buttons are correctly displayed
        }

        // Event Delegation for options
        optionsListEl.addEventListener('click', (event) => {
            if (event.target && event.target.matches('li.option-item')) {
                // Check if the question has already been answered
                if (userAnswers[currentQuestionIndex] === null) {
                    const selectedIndex = parseInt(event.target.dataset.index, 10);
                    selectOption(selectedIndex);
                }
            }
        });

        prevBtn.addEventListener('click', goToPreviousQuestion);
        nextBtn.addEventListener('click', goToNextQuestion);
        hintBtn.addEventListener('click', showHint);
        
        // 初始加载
        loadQuestion();
      </script>
    </body>
    </html>
  